A307453 a(n) is the least prime p for which the continued fraction expansion of sqrt(p) has exactly n consecutive 1's starting at position 2.

2, 3, 31, 7, 13, 3797, 5273, 4987, 90371, 79873, 2081, 111301, 1258027, 5325101, 12564317, 9477889, 47370431, 709669249, 1529640443, 2196104969, 392143681, 8216809361, 30739072339, 200758317433, 370949963971, 161356959383, 1788677860531, 7049166342469, 4484287435283, 3690992602753

Offset: 0

Author

Michel Marcus, Apr 09 2019

Keywords

• nonn

Examples

			For p = 2,  we have [1; 2, ...]; see A040000.
For p = 3,  we have [1; 1, 2, ...]; see A040001.
For p = 31, we have [5; 1, 1, 3, ...]; see A010129.
For p = 7,  we have [2; 1, 1, 1, 4, ...]; see A010121.
		

Links

• Piotr Miska, Maciej Ulas, On consecutive 1's in continued fractions expansions of square roots of prime numbers, arXiv:1904.03404 [math.NT], 2019. See Table 1 p. 15.
• Index entries for continued fractions for constants

Crossrefs

Cf. A040000, A040001, A010121, A010129.

Programs

PARI

isok(p, n) = {my(c=contfrac(sqrt(p)));  for (k=2, n+1, if (c[k] != 1, return (0));); return(c[n+2] !=  1);}
a(n) = {my(p=2); while (! isok(p, n), p = nextprime(p+1)); p;}

Formula

Limit_{n->infinity} (sqrt(a(n)) - floor(sqrt(a(n)))) = A094214. - Daniel Suteu, Apr 09 2019

Extensions

a(21)-a(29) from Daniel Suteu, Apr 09 2019

a(0) added by Chai Wah Wu, Apr 09 2019